Probability of choosing a biased coin $C$ which has probability $3/15$ of getting heads, assuming we got head...
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Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?
My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
$$frac1{15} cdot frac12 = frac1{30}$$
probability discrete-mathematics conditional-probability
edited Dec 24 '18 at 7:49
Asaf Karagila♦
303k32429760
asked Dec 24 '18 at 4:50
hussain sagarhussain sagar
806
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add a comment |
$begingroup$
Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?
My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
$$frac1{15} cdot frac12 = frac1{30}$$
probability discrete-mathematics conditional-probability
edited Dec 24 '18 at 7:49
Asaf Karagila♦
303k32429760
asked Dec 24 '18 at 4:50
hussain sagarhussain sagar
806
$endgroup$
add a comment |
$begingroup$
Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?
My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
$$frac1{15} cdot frac12 = frac1{30}$$
probability discrete-mathematics conditional-probability
edited Dec 24 '18 at 7:49
Asaf Karagila♦
303k32429760
asked Dec 24 '18 at 4:50
hussain sagarhussain sagar
806
$endgroup$
Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?
My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
$$frac1{15} cdot frac12 = frac1{30}$$
probability discrete-mathematics conditional-probability
probability discrete-mathematics conditional-probability
edited Dec 24 '18 at 7:49
Asaf Karagila♦
303k32429760
asked Dec 24 '18 at 4:50
hussain sagarhussain sagar
806
edited Dec 24 '18 at 7:49
Asaf Karagila♦
303k32429760
asked Dec 24 '18 at 4:50
hussain sagarhussain sagar
806
edited Dec 24 '18 at 7:49
Asaf Karagila♦
303k32429760
edited Dec 24 '18 at 7:49
Asaf Karagila♦
303k32429760
edited Dec 24 '18 at 7:49
Asaf Karagila♦
303k32429760
303k32429760
asked Dec 24 '18 at 4:50
hussain sagarhussain sagar
806
asked Dec 24 '18 at 4:50
hussain sagarhussain sagar
806
asked Dec 24 '18 at 4:50
hussain sagarhussain sagar
806
806
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989bgt6989b
33.9k22455
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Thank you, this makes a lot of sense.
$endgroup$
– hussain sagar
Dec 24 '18 at 4:57
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@hussainsagar you are welcome
$endgroup$
– gt6989b
Dec 24 '18 at 4:58
add a comment |
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What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered Dec 24 '18 at 4:55
Siong Thye GohSiong Thye Goh
100k1466117
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
Thank you, this makes a lot of sense.
$endgroup$
– hussain sagar
Dec 24 '18 at 4:57
$begingroup$
@hussainsagar you are welcome
$endgroup$
– gt6989b
Dec 24 '18 at 4:58
add a comment |
$begingroup$
An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
Thank you, this makes a lot of sense.
$endgroup$
– hussain sagar
Dec 24 '18 at 4:57
$begingroup$
@hussainsagar you are welcome
$endgroup$
– gt6989b
Dec 24 '18 at 4:58
add a comment |
$begingroup$
An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989bgt6989b
33.9k22455
$endgroup$
An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989bgt6989b
33.9k22455
answered Dec 24 '18 at 4:55
gt6989bgt6989b
33.9k22455
answered Dec 24 '18 at 4:55
gt6989bgt6989b
33.9k22455
answered Dec 24 '18 at 4:55
gt6989bgt6989b
33.9k22455
33.9k22455
$begingroup$
Thank you, this makes a lot of sense.
$endgroup$
– hussain sagar
Dec 24 '18 at 4:57
$begingroup$
@hussainsagar you are welcome
$endgroup$
– gt6989b
Dec 24 '18 at 4:58
add a comment |
$begingroup$
Thank you, this makes a lot of sense.
$endgroup$
– hussain sagar
Dec 24 '18 at 4:57
$begingroup$
@hussainsagar you are welcome
$endgroup$
– gt6989b
Dec 24 '18 at 4:58
$begingroup$
Thank you, this makes a lot of sense.
$endgroup$
– hussain sagar
Dec 24 '18 at 4:57
$begingroup$
Thank you, this makes a lot of sense.
$endgroup$
– hussain sagar
Dec 24 '18 at 4:57
$begingroup$
@hussainsagar you are welcome
$endgroup$
– gt6989b
Dec 24 '18 at 4:58
$begingroup$
@hussainsagar you are welcome
$endgroup$
– gt6989b
Dec 24 '18 at 4:58
add a comment |
$begingroup$
What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered Dec 24 '18 at 4:55
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered Dec 24 '18 at 4:55
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered Dec 24 '18 at 4:55
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered Dec 24 '18 at 4:55
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 24 '18 at 4:55
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 24 '18 at 4:55
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 24 '18 at 4:55
Siong Thye GohSiong Thye Goh
100k1466117
100k1466117
add a comment |
add a comment |
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